Mechanical Engineering - Machine Design - Discussion
Discussion Forum : Machine Design - Section 9 (Q.No. 5)
5.
The permissible stress in the fillet weld is 100 N/mm2. The fillet weld has equal leg lengths of 15 mm each. The allowable shearing load on the weldment per cm length of the weld is
Discussion:
3 comments Page 1 of 1.
Harshdeepsinh said:
6 years ago
No, @Jaypatel.
Because if you want to convert this quantity into cm then you have to divide this1060.5 N per mm by 10. They have to give the value of permissible stress 1000N/mm2. Then it should be 10.69KN.
Because if you want to convert this quantity into cm then you have to divide this1060.5 N per mm by 10. They have to give the value of permissible stress 1000N/mm2. Then it should be 10.69KN.
Jay patel said:
9 years ago
= 0.707 * leg length * weld length * permissible limit,
= 0.707 * 15 * 1 (mm) * 100,
= 1060.5 N per mm,
= 10605 N per cm,
= 10.60 kN
= 0.707 * 15 * 1 (mm) * 100,
= 1060.5 N per mm,
= 10605 N per cm,
= 10.60 kN
(1)
Imran said:
10 years ago
Please send me formula used.
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